N的最小倍数，其二进制数表示形式中的N个数字正好

给定正整数N ，任务是在其二进制数表示形式中找到N个具有N个数字的最小倍数。
例子：

Input: N = 3 
Output: 6 
Explanation: 
6 is the smallest multiple of 3 and has length also 3(110) in binary.
Input: N = 5 
Output: 20 
Explanation: 
6 is the smallest multiple of 5 and has length also 5(10100) in binary. 
 

方法：想法是观察。

• 如果我们仔细观察，一个序列将形成为1、2、6、8、20 …
• 系列中的第N个术语是：

$N*\lceil 2^\frac{N-1}{N} \rceil$
 

• 因此，将数字N作为输入并实现以上公式。

下面是上述方法的实现：

C++

// C++ program to find smallest
// multiple of n with exactly N
// digits in Binary number System.
 
#include 
#include 
using namespace std;
 
// Function to find smallest multiple
// of n with exactly n digits
// in Binary number representation.
void smallestNumber(int N)
{
    cout << N * ceil(pow(2,
                         (N - 1))
                     / N);
}
 
// Driver code
int main()
{
    int N = 3;
    smallestNumber(N);
 
    return 0;
}

Java

// Java program to find smallest
// multiple of n with exactly N
// digits in Binary Number System.
class GFG{
 
// Function to find smallest
// multiple of n with exactly N
// digits in Binary Number System.
static void smallestNumber(int N)
{
    System.out.print(N * Math.ceil
                        (Math.pow(2, (N - 1)) / N));
}
 
// Driver code
public static void main(String[] args)
{
    int N = 3;
     
    smallestNumber(N);
}
}
 
// This code is contributed by shubham

Python3

# Python3 program to find smallest
# multiple of n with exactly N
# digits in Binary number System.
from math import ceil
 
# Function to find smallest multiple
# of n with exactly n digits
# in Binary number representation.
def smallestNumber(N):
    print(N * ceil(pow(2, (N - 1)) / N))
 
# Driver code
N = 3
smallestNumber(N)
 
# This code is contributed by Mohit Kumar

C#

// C# program to find smallest
// multiple of n with exactly N
// digits in Binary Number System.
using System;
 
class GFG{
 
// Function to find smallest
// multiple of n with exactly N
// digits in Binary Number System.
static void smallestNumber(int N)
{
    Console.Write(N * Math.Ceiling(
                      Math.Pow(2, (N - 1)) / N));
}
     
// Driver code
public static void Main(string[] args)
{
    int N = 3;
         
    smallestNumber(N);
}
}
 
// This code is contributed by AnkitRai01

Javascript

// Javascript program to find smallest
// multiple of n with exactly N
// digits in Binary number System.
 
// Function to find smallest multiple
// of n with exactly n digits
// in Binary number representation.
function smallestNumber(N)
{
    document.write(N * parseInt(Math.ceil(Math.pow(2,
                                 (N - 1))
                             / N)));
}
 
// Driver code
let N = 3;
smallestNumber(N);
 
// This code is contributed by rishavmahato348.

输出：

6